Chapter 9, Problem 65A

Interpretation Introduction

Interpretation: The percent yield for the given reaction under given conditions needs to be calculated.

Concept introduction: In a chemical reaction, the amount of product formed in the relation to the amount of reactant consumed is term as yield. Yield is generally expressed in percentage.

Answer to Problem 65A

The percent yield for the given reaction is $73.38\text{ \%}$ .

Explanation of Solution

Given:

  ${\text{KClO}}_{\text{3}}\text{(s) }\to {\text{ KCl(s) + O}}_{\text{2}}\text{(g)}$

The mass of sample decomposed is, ${\text{KClO}}_{\text{3}}=\text{ 13}\text{.97 g}$ .

The mass of $\text{KCl}$ is $\text{6}\text{.23 g}$ .

The given reaction is not balanced, as the number of oxygen on reactant side is 3 and on the product side is 2. So, in order to balance the reaction coefficient 2 is written before KClO₃ on the reactant side and 2 before KCl and 3 before O₂ on the product side. Thus, the balanced reaction is:

  ${\text{2KClO}}_{\text{3}}\text{(s) }\to {\text{ 2KCl(s) + 3O}}_{\text{2}}\text{(g)}$

The number of moles of ${\text{KClO}}_{\text{3}}$ is calculated as:

  $\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

The molar mass of ${\text{KClO}}_{\text{3}}$ is 122.55 g/mol. So, the number of moles of ${\text{KClO}}_{\text{3}}$ is:

  $\begin{array}{l}{\text{n}}_{{\text{KClO}}_{\text{3}}}=\frac{13.97\text{ g}}{\text{122}\text{.55 g/mol}}\\ {\text{n}}_{{\text{KClO}}_{\text{3}}}=0.114\text{ mol}\end{array}$

From balanced reaction, the mole ratio of ${\text{KClO}}_{\text{3}}$ and $\text{KCl}$ is 2:2 that is 1:1 that means 1 mole of ${\text{KClO}}_{\text{3}}$ produces 1 mole of $\text{KCl}$ . So, the number of moles of $\text{KCl}$ produced from 0.114 mol of ${\text{KClO}}_{\text{3}}$ is 0.114 mol.

Now, the mass of produced $\text{KCl}$ is calculated as:

  $\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

Molar mass of $\text{KCl}$ is 74.55 g/mol. So,

  $\begin{array}{l}\text{0}\text{.114 mol}=\frac{\text{Mass}}{\text{74}\text{.55 g/mol}}\\ \text{Mass = 0}\text{.114 mol}\times \text{74}\text{.55 g/mol}\\ \text{Mass = }8.49\text{ g}\end{array}$

Now, the percent yield is calculated using formula:

  $\text{Percent}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100$

Substituting the values:

  $\begin{array}{l}\text{Percent yield}=\frac{\text{6}\text{.23 g}}{\text{8}\text{.49 g}}\times 100\text{ \%}\\ \text{Percent yield}=73.38\text{ \%}\end{array}$

Hence, the percent yield for the given reaction is $73.38\text{ \%}$ .